Question

An uncharged capacitor of capacitance 100 microFarad is connected to a battery of emf 20V at t=0 through a resistance 10ohm ,then time at which the rate of energy stored in capacitor is maximum:
(A)(4 ln 2)ms
(B)(2 ln 2)ms
(C)(ln 2)ms
(D)(3 ln 2)ms

Answer 1

Explanation:

see the adjoining attachment..

Thank you..

Answer 2

Given that,

Capacitance of uncharged capacitor = 100 μF

Emf = 20 V at t = 0

Resistance = 10 ohm

We need to calculate the minimum time

Using formula of time

$t=5\times\tau$

$t=5\times R\times C$

Put the value into the formula

$t=5\times10\times100\times10^{-6}$

$t=0.005\ sec$

We need to calculate the stored energy

Using formula of stored energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times100\times10^{-6}\times20^2$

$E=0.02\ J$

We need to calculate the time at which the rate of energy stored in capacitor is maximum

Using formula of current

$i=\dfrac{V}{R}e^{\frac{-t}{RC}}$

$C\dfrac{dv}{dt}=\dfrac{V}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$100\times10^{-6}\times\dfrac{0.02}{0.005}=\dfrac{20}{10}e^{\frac{-t}{10\times100\times10^{-6}}}$

$0.0004=2e^{\frac{-t}{10\times100\times10^{-6}}}$

$ln(0.0004)=2\times\dfrac{-t}{10^{-3}}$

After solving,

$t= ln2\ ms$

Hence, The time is $ln2\ ms$

(C) is correct option.